My self-similarity function is defined by : R(t) = $$ \int_{-\infty}^\infty \mathrm y(x+t)y^\ast(x)\,\mathrm{d}x $$
which is apparently equal to R(t) = $$ \int_{-\infty}^\infty \mathrm Y(f)Y^\ast(f){e}^{i2{\pi}tf}\,\mathrm{d}f $$
Y is the fourier transform of y.
I tried to find it using inverse fourier transform but I couldn't find out how.
Thank you for helping !
The easiest way to see the equality is using Parseval's Theorem: In general, $$ \int f g^\ast\, dx = \int F G^\ast\, df $$ where F and G are the Fourier transforms of f and g.
Combine this with the fact that, if $Y(f)$ is the transform of $y(x)$, then the transform of $y(x+t)$ is $Y(f) e^{2\pi i\, tf}$ (depending on the normalization of the Fourier transform) and you are done.